Difference between revisions of "1989 AIME Problems/Problem 5"
Matrix math (talk | contribs) (→Solution 2) |
Matrix math (talk | contribs) (→Solution 2) |
||
Line 8: | Line 8: | ||
=== Solution 2 === | === Solution 2 === | ||
− | Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting 3 heads, we need to find <math>{5\choose3}\frac{(h)^3(t)^2}{(h + t)^5} =10\cdot\frac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be <math>0</math>). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>. | + | Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting <math>3</math> heads, we need to find <math>{5\choose3}\frac{(h)^3(t)^2}{(h + t)^5} =10\cdot\frac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be <math>0</math>). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>. |
== See also == | == See also == |
Revision as of 09:43, 25 April 2017
Problem
When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to and is the same as that of getting heads exactly twice. Let , in lowest terms, be the probability that the coin comes up heads in exactly out of flips. Find .
Solution
Solution 1
Denote the probability of getting a heads in one flip of the biased coin as . Based upon the problem, note that . After canceling out terms, we get , so . The answer we are looking for is , so .
Solution 2
Denote the probability of getting a heads in one flip of the biased coins as and the probability of getting a tails as . Based upon the problem, note that . After cancelling out terms, we end up with . To find the probability getting heads, we need to find (recall that cannot be ). The result after simplifying is , so .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.